Problem 77 - Calculus 3

Question

Find the equation of the tangent line to the curve \(\mathbf{r}(t)=\left\langle e^{t}, e^{-t}, 0\right\rangle\) at \( t=0\)

Step-by-Step Solution

Answer

The tangent line is \( \mathbf{L}(t) = \langle 1 + t, 1 - t, 0 \rangle \).

Step 1: Differentiate the Vector Function

Find the derivative of the vector function \( \mathbf{r}(t) = \langle e^t, e^{-t}, 0 \rangle \) with respect to \( t \). The derivative, \( \mathbf{r}'(t) \), gives the tangent vector at any point \( t \). For each component, differentiate separately:- \( \frac{d}{dt} e^t = e^t \)- \( \frac{d}{dt} e^{-t} = -e^{-t} \)Thus, the derivative is \( \mathbf{r}'(t) = \langle e^t, -e^{-t}, 0 \rangle \).

Step 2: Evaluate the Derivative at t = 0

Substitute \( t = 0 \) into \( \mathbf{r}'(t) \) to find the tangent vector at the given point. Calculate each component:- \( e^0 = 1 \)- \( -e^0 = -1 \)The tangent vector at \( t = 0 \) is \( \mathbf{r}'(0) = \langle 1, -1, 0 \rangle \).

Step 3: Find the Point on the Curve at t = 0

Evaluate the original vector function \( \mathbf{r}(t) \) at \( t = 0 \) to find the point on the curve:- \( \mathbf{r}(0) = \langle e^0, e^0, 0 \rangle = \langle 1, 1, 0 \rangle \).

Step 4: Formulate the Tangent Line Equation

The equation of the tangent line can be written as \( \mathbf{L}(t) = \mathbf{r}(0) + t \cdot \mathbf{r}'(0) \), where \( \mathbf{r}(0) \) is the point on the curve and \( \mathbf{r}'(0) \) is the tangent vector. Substitute the values:\[ \mathbf{L}(t) = \langle 1, 1, 0 \rangle + t \cdot \langle 1, -1, 0 \rangle \]Simplify:\[ \mathbf{L}(t) = \langle 1 + t, 1 - t, 0 \rangle \].

Key Concepts

Vector Function

A vector function is a function that takes one or more variables and returns a vector as its output. In the context of curves in three-dimensional space, a vector function often represents the position vector of a point on the curve. For example, the vector function given in the exercise is \( \mathbf{r}(t) = \langle e^t, e^{-t}, 0 \rangle \). This vector function describes a curve in 3D space where each component of the vector is a mathematical expression depending on the parameter \( t \). Here:

• \( e^t \) is the x-coordinate.
• \( e^{-t} \) is the y-coordinate.
• The z-coordinate is always 0.

These components control the position of a point on the curve as \( t \) changes. Understanding vector functions is crucial because they provide a neat and powerful way to describe curves and surfaces in multidimensional spaces.

Derivative

The derivative of a vector function is a key concept when analyzing curves. It indicates how the curve changes or reacts as the parameter changes. For the vector function \( \mathbf{r}(t) = \langle e^t, e^{-t}, 0 \rangle \), each component function needs to be differentiated with respect to \( t \).Let's break it down:

• The derivative of \( e^t \) with respect to \( t \) is \( e^t \).
• The derivative of \( e^{-t} \) is \(-e^{-t} \).
• The derivative of the constant \( 0 \) is simply \( 0 \).

Thus, the derivative is \( \mathbf{r}'(t) = \langle e^t, -e^{-t}, 0 \rangle \). This new vector, \( \mathbf{r}'(t) \), gives us the tangent vector of the curve, which is a vector that points in the direction in which the curve is heading at any point \( t \). By understanding the derivative, we can better understand the "behavior" of the curve over time.

Tangent Vector

The tangent vector is crucial for finding the tangent line to a curve. It represents the instantaneous direction and speed of movement along the curve. When taking the derivative of the vector function, as seen before, you get this vector.In the exercise, the derivative \( \mathbf{r}'(t) = \langle e^t, -e^{-t}, 0 \rangle \) is evaluated at \( t = 0 \) to give the tangent vector \( \mathbf{r}'(0) = \langle 1, -1, 0 \rangle \). This specific vector captures:

• The curve's direction at the point of tangency.
• The rate at which the point moves along the x and y directions (no change in the z-direction).

Using this tangent vector, we can construct a parametric equation of the tangent line that describes the line tangent to the curve at the specific point.

Parametric Equations

In calculus, parametric equations provide a comprehensive way to describe a curve through a set of equations by expressing the coordinates of points on the curve as functions of a parameter, \( t \). They offer flexibility and ease when describing complex curves and surfaces.For the tangent line in the exercise, the parametric equation is given by \( \mathbf{L}(t) = \mathbf{r}(0) + t \cdot \mathbf{r}'(0) \), where:

• \( \mathbf{r}(0) \) is the point on the curve at \( t=0 \) which results in \( \langle 1, 1, 0 \rangle \).
• \( \mathbf{r}'(0) \) is the tangent vector \( \langle 1, -1, 0 \rangle \).

This leads to the equation \( \mathbf{L}(t) = \langle 1 + t, 1 - t, 0 \rangle \), showing that as \( t \) varies, the "tangent line" travels along a straight path, maintaining the direction and speed captured by the tangent vector.